Ampère's Law
Introduced| English | Meaning |
|---|---|
| Ampère's law/ˈæmpeəz lɔː/ | In magnetostatics in vacuum, the circulation of the magnetic field around a closed path equals the enclosed current multiplied by the magnetic constant. |
| Amperian loop/æmˈpɪərɪən luːp/ | A chosen closed integration path used when applying Ampère's law. |
Gauss's law has a magnetic cousin — a shortcut for B
- Adding up Biot–Savart pieces for a whole coil is hard work.
- When the field is symmetric, there is a one-line shortcut.
- Ampère's law 安培定律 links the field around a loop to the current through it.
- It plays the same role for $B$ that Gauss's law plays for $E$.
Ampère's law plays the same role for the magnetic field that ____ plays for the electric field.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Both are symmetry shortcuts: Gauss for E, Ampère for B.
The law itself
- Around any closed loop: $\oint \vec B \cdot d\vec\ell = \mu_0 I_{\text{enc}}$.
- The left side sums the field along the loop; the right is the enclosed current.
- Only the current threading the loop counts.
- Choose the loop cleverly and $B$ pops right out.
In Ampère's law, the field summed around a loop equals:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
$\oint \vec B \cdot d\vec\ell = \mu_0 I_{\text{enc}}$.
Only enclosed current matters
- A current outside the loop adds nothing to the line integral.
- Its field goes around one side of the loop and back on the other — it cancels.
- So the right side counts only $I_{\text{enc}}$, the current passing through.
- Same idea as Gauss's law, with current in place of charge.
A current outside an Amperian loop contributes nothing to the line integral of B.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Only the enclosed current counts; outside current cancels around the loop.
Pick an Amperian loop with the symmetry
- Straight wire → a circular Amperian loop 安培环路 around it.
- Solenoid → a rectangular loop through its side.
- Choose it so $B$ is constant and along (or across) each part of the loop.
- Then $\oint \vec B \cdot d\vec\ell$ becomes a simple $B \times (\text{length})$.
Which current is enclosed?
Ampère's law uses only the current threading the loop. Sort each wire.
The closed path you integrate B around is called an ____ loop.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
It is an Amperian loop, chosen to match the symmetry.
Select all true statements about Ampère's law.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Always true, best with symmetry, enclosed current only — around a closed loop.
Fast, clean results
- Around a straight wire: $B = \dfrac{\mu_0 I}{2\pi r}$ — recovered in one line.
- Inside a long solenoid: $B = \mu_0 n I$, where $n$ is turns per metre.
- Inside a toroid, the field is neatly confined to the ring.
- Each drops out of Ampère's law with almost no algebra.
Ampère's law gives the field inside a long solenoid as:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Inside a solenoid $B = \mu_0 n I$ ($n$ = turns per metre).
Use a circular Amperian loop of radius $r$ around a straight wire carrying $I$.
- Symmetry: $\oint B\,d\ell = B(2\pi r)$.
- Set equal to $\mu_0 I$: $B = \dfrac{\mu_0 I}{2\pi r}$ — the straight-wire result.
Like Gauss's law, Ampère's law is always true but only useful when symmetry makes $B$ constant along your loop. And only the enclosed current counts — a nearby wire outside the loop changes $B$ but adds nothing to $\oint \vec B \cdot d\vec\ell$.
Ampère's law says $\oint \vec B \cdot d\vec\ell = \mu_0 I_{\text{enc}}$ — the field around a loop depends only on the enclosed current. Pick an Amperian loop matching the symmetry to read $B$ off in one step (straight wire $\mu_0 I/2\pi r$, solenoid $\mu_0 n I$).
Write from a solenoid field of 4 pi times 10 to the minus 4 tesla
- A long solenoid has 500 turns per metre and carries 2 amperes.
- Find the field inside. Say the ideal field outside.
A long solenoid has 500 turns per metre and carries 2 amperes. What is the field inside, and what is the ideal field outside?
n I is 1000 amperes per metre. mu0 times that current density is 4 pi times 10 to the minus 4 tesla.
What is the solenoid field?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
mu0 times 500 times 2 is 4 pi times 10 to the minus 4 tesla inside an ideal solenoid.